Factoring Distributive Law Special Forms Binomials Quadratic Formula 1. Distributive Law a(b + c) = ab + ac Factor 2x2 + 4x = 2(x2) + 2(2x) = 2(x2 + 2x) 2x(x + 2) 2x2 + 4x = 2x(x + 2) Example Each term has a common factor of 2. Factor out the 2. And use DL to place it out in front. Each term also has a common X. Use the DL to

factor it out front. Since the terms have nothing else in common, you are done. 2. Special Forms x2 a2 = (x a)(x + a) x3 a3 = (x a)(x2 + ax + a2) x3 + a3 = (x + a)(x2 ax + a2) These are the basic special forms that you must remember! 2. Special Forms x2 a2 = (x a)(x + a) 2 xTo find 4 a, take the square root of 4. So, a = = x2 22 2. 2)(x + 2)the square root of4 7. So, a =

= To(xfind a, take Since x2 7 it doesnt reduce, is the number. = = (x 7 )(x + 7 ) = . 7 7 2. Special Forms x2 a2 = (x a)(x + a) 2 xTo find 28 a, take the square root of 28. 2. Special Forms

x2 a2 = (x a)(x + a) To find the square root28 of 28. Do a factor tree on 28. 2 2 and 7 are prime. 14 2 7 2. Special Forms x2 a2 = (x a)(x + a) Now, take28 the square root by boxing 2 of the same number. 2 14 Place the number in each box out in front with a

2 7 power equal to the number of boxes. All leftovers remain under the radical sign. 28 = 2 7 2. Special Forms x2 a2 = (x a)(x + a) 2 xTo find 28 a, take the square root of 28. Now that we have 7 ) the answer using the = (x2 2 , 7we )(xcan + 2write a= 7 special form.

2. Special Forms x3 a3 = (x a)(x2 + ax + a2) 3 xTo find 27 a, take the cube root of 27. 2. Special Forms x3 a3 = (x a)(x2 + ax + a2) To find3the 27cube root of 27. Do a factor tree on 27. 3 3 is prime. 9 3 3 2. Special Forms

x3 a3 = (x a)(x2 + ax + a2) 3 27 Now, take the cube root by boxing 3 of the same number. 3 9 Place the number in each box out in front with a power equal to the number of boxes. All 3 3 leftovers remain under the radical sign. 3 27 = 3 2. Special Forms x3 a3 = (x a)(x2 + ax + a2) 3

xTo find 27 a, take the cube root of 27. Now that we have 2 2 = (x 3)(x + 3x + 3 ) answer using the special a = 3, we can write the form. = (x 3)(x2 + 3x + 9) 2. Special Forms x3 + a3 = (x + a)(x2 ax + a2)

3 xTo +find 64 a, take the cube root of 64 as before. a=4 Use the special form and note that the only difference is the sign in the middle of each part. x3 + 64 = (x + 4)(x2 4x + 16) 2. Special Forms x3 + a3 = (x + a)(x2 ax + a2) All positive x3 a3 = (x a)(x2 + ax + a2) First sign matches Alternates x3 + a3 = (x + a)(x2 ax + a2)

3. Binomial Products (a b)2 = a2 2ab + b2 A perfect square always has this form EXAMPLES: (x + 3)2 = (x)2 + 2(3)(x) + (3)2 = x2 + 6x + 9 (x 5)2 = (x)2 + 2(-5)(x) + (-5)2 = x2 10x + 25 (x + )2 = (x)2 + 2()(x) + ()2 = x2 + x + 1/16 3. Binomial Products (a b)2 = a2 2ab + b2 Is it a perfect square? x2 + 5x + 9 (x)2 + 5x + (3)2 Write each end as a square (3)(x) Find the product of the bases 2(3)(x) And double it! 6x Is this whats in the middle? If it doesnt match, it is not a perfect square move on.

3. Binomial Products (a b)2 = a2 2ab + b2 Is it a perfect square? x2 + 3x + 4 x2 5x + 25 9x2 + 12x + 4 YES x2 12x + 36 YES 2(2)(x) = 4x 2(5)(x) = 10x 2(3x)(2) = 12x 2(6)(x) = 12x NO NO 3. Binomial Products (a b)2 = a2 2ab + b2

If it is a perfect square, write the answer 9x2 + 12x + 4 2(3x)(2) = 12x YES 9x2 + 12x + 4 = (3x + 2)2 x2 12x + 36 YES x2 12x + 36 = (x 6)2 2(6)(x) = 12x 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab To set up a table to factor a quadratic, you need to identify the factors of the constant term (last number) that add to give you the number in front of the x in the middle. x2 + (a + b)x + ab

3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 3x + 2 a b a+b What factors of 2 (the last number) will add up to give you -3 (the number in front of x)? 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 3x + 2 (2)(1) = 2 but, 2 + 1 = 3, not -3. a b

a+b 2 1 3 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 3x + 2 a b a+b 2 1

3 When you have the right size the wrong -2 (3) -1 but -3 sign (+ instead of -), change the signs in the a and b columns. 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 3x + 2 Now the numbers work! (-2)(-1) = 2 (-2) + (-1) = -3 a b a+b

2 1 3 -2 -1 -3 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab x2 3x + 2 = (x 2)(x 1) a b

a+b 2 1 3 -2 -1 -3 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab x2 3x + 2 = (x 2)(x 1) a b

a+b 2 1 3 -2 -1 -3 The signs of the numbers go in the factors. 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab x2 3x + 2 = (x 2)(x 1) Done!

a b a+b 2 1 3 -2 -1 -3 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 + 4x 3

a b a+b What factors of -3 will add up to give you 4? 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 + 4x 3 a b a+b 1 -3

-2 3 -1 2 None of them work?! Now what? 3. Binomial Products (x + a)(x + b) = x2 + (a + b)x + ab Factor x2 + 4x 3 It doesnt work. Move on to something else. Done! a

b a+b 2 1 3 -2 -1 -3 3. Binomial Products Vietas Method To set up a table to factor a quadratic, you need to identify the factors of a product of the leading (first) and constant (last) terms

that add to give you the number in front of the x in the middle. (ax + b)(cx + d) = acx2 + (ad + bc)x + bd 3. Binomial Products Vietas Method (ax + b)(cx + d) = acx2 + (ad + bc)x + bd parts of parts of leading term constant The product of the leading and constant terms includes all of the pieces (abcd). 3. Binomial Products Vietas Method Factor 3x2 8x + 4 a

b a+b What factors of (3)(4)=12 will add to give you -8 (the coefficient of x)? 3. Binomial Products Vietas Method Factor 3x2 8x + 4 (1)(12)=12 (2)(6)=12 (3)(4)=12 Dont forget negatives (-1)(-12)=12 (-2)(-6)=12 (-3)(-4)=12 a

b a+b 1 12 13 2 6 8 3 4 7

-1 -12 -13 -2 -6 -8 -3 -4 -7 3. Binomial Products Vietas Method Factor 3x2 8x + 4

(-2) + (-6) = -8 a b a+b 1 12 13 2 6 8 3

4 7 -1 -12 -13 -2 -6 -8 -3 -4 -7

These are the factors we want! 3. Binomial Products Vietas Method Factor 3x2 8x + 4 Split the middle term a b a+b 1 12 13

2 6 8 3 4 7 3x2 2x 6x-1+ 4-12 -13 -2 -6 -8 It doesnt matter what order you put them in! -3 -4 -7

3. Binomial Products Vietas Method Factor 3x2 8x + 4 a b a+b 1 12 13 2 6 8

3 4 7 3x2 2x 6x-1+ 4-12 -13 -2 -6 -8 It doesnt matter what order you put them in! -3 -4 -7 3. Binomial Products Vietas Method Factor 3x2 8x + 4

a b a+b Group the first and last pairs1 of 12 terms13 2 6 (3x2 2x) + (6x 3 + 44) 8 7 -1 -12

-13 -3 -4 -7 signs go with the numbers! -2 between -6 the-8 If there is a -, you must insert a + sign parentheses. 3. Binomial Products Vietas Method Factor 3x2 8x + 4 a b

a+b 12 13 Use distributive law on each1 group. 2 6 x(3x 2) 2(3x 3 2) 4 8 7 -1 -12 -13

-3 -4 -7 Watch your signs! -2 -6 -8 If there is a - beginning a group, take it out. 3. Binomial Products Vietas Method Factor 3x2 8x + 4 Look for shared group a b

a+b 1 12 13 2 6 8 3 4 7 x(3x 2) 2(3x

2) -1 -12 -13 -2 -6 -8 -3 -4 -7 If there is a shared group, you did it right! 3. Binomial Products Vietas Method

Factor 3x2 8x + 4 a b a+b 12 group 13 Use Distributive Law on the1shared 2 6 8 (3x 2)(x 32) 4 7 -1

-12 -13 -2 -6 -8 -3 -4 -7 3. Binomial Products Vietas Method Factor 3x2 8x + 4

a b a+b 1 12 13 2 6 8 3 4

7 = (3x 2)(x-1 2)-12 -13 -2 -6 -8 -3 -4 -7 DONE! 4. Quadratic Formula

Given ax2 + bx + c, find r1 and r2 by r1,2 = , one uses the + sign and the other uses the sign Then ax2 + bx + c always factors into ax2 + bx + c = a(x r1)(x r2) 4. Quadratic Formula Why dont we just use this all of the time? Its work! Plug into the formula and write down the results. 4. Quadratic Formula Factor 4x2 3x 7 r1,2 = = = = = = , -1 = , -1

4x2 3x 7 = 4(x )(x + 1) = (4x 7)(x + 1) Factoring Use the Methods in the following Order 1. Distributive Law a(b + c) = ab + ac 2. Special Forms x2 a2 = (x a)(x + a) x3 a3 = (x a)(x2 + ax + a2) x3 + a3 = (x + a)(x2 ax + a2) 3. Binomial Products

(x a)2 = x2 2ax + a2 (x + a)(x + b) ax2 + bx + c 4. Quadratic Formula